Question

Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ?

Answer 1

Answer: The weight of X is $33\dfrac{1}{3}\%$ of weight of mixture.

Step-by-step explanation:

Since we have given that

Percentage of seed mixture X for ryegrass = 40%

Percentage of seed mixture Y for ryegrass = 25%

If a mixture of X and Y contains 30 percent ryegrass,

Let total seed mixture be 100

So, for seed X = x

For seed Y = 100-x

So, According to question,

$0.4x+0.25(100-x)=30\\\\0.4x+25-0.25x=30\\\\0.15x=30-25\\\\0.15x=5\\\\x=\dfrac{5}{0.15}\\\\x=\dfrac{100}{3}$

So, weight of mixture X is given by

$\dfrac{\text{Weight of X}}{\text{Weight of mixture}}\times 100\\\\=\dfrac{\dfrac{100}{3}}{100}\times 100\\\\=\dfrac{100}{3}\%\\\\=33\dfrac{1}{3}\%$

Hence, the weight of X is $33\dfrac{1}{3}\%$ of weight of mixture.